problem 32

/*
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
*/
#include <iostream>
#include<vector>
#include<string>
#include<set>
#include<algorithm>
using namespace std;


int main() {
  int x , y , z;
  int sum = 0;
  set<int> total;
  vector<int> numbers = {1, 2, 3, 4, 5, 6, 7, 8, 9};
  do{
     for(int i = 1; i < 9; i++){
       for(int j = 1; j < 9 - i; j++){
         int rem_len = 9 - i - j;
         if(rem_len < i || rem_len  < j){
           break;
         }
         int pos = 0;
         int first_product = 0;
         for(x = 0; x < i; x++){
           first_product *= 10;
           first_product += numbers[pos++];
         } 
         int second_product = 0;
         for(y = 0; y < j; y++){
             second_product *= 10;
             second_product += numbers[pos++];
         }
         int last_product = 0;
         for(z = 0; z < rem_len; z++){
           last_product *= 10;
           last_product += numbers[pos++];
         }
         if(first_product * second_product  == last_product){
           total.insert(last_product);
         }
       }
     }
  }
 while(std::next_permutation(numbers.begin() , numbers.end()));
 for(int n : total){
   sum += n;
 }
 cout << sum <<endl;
 return 0;
}